Problem 2263

To find all triplets of positive integers $(a, m, n)$ such that $a^m + 1 \mid (a + 1)^n$, we need to analyze the divisibility condition $a^m + 1 \mid (a + 1)^n$. This condition suggests that $(a + 1)^n = k(a^m + 1)$ for some integer $k$.

**Step 1: Analyze cases where $m = 1$:**

If $m = 1$, then the divisibility condition becomes:
$$a + 1 \mid (a + 1)^n$$
which is true for all $n$ since $(a + 1)$ clearly divides $(a + 1)^n$. Thus, for $m = 1$, any triplet $(a, 1, n)$ satisfies the condition.

**Step 2: Analyze cases where $a = 1$:**

If $a = 1$, the condition becomes:
$$1^m + 1 = 2 \mid (1 + 1)^n = 2^n$$
This is true for all $m$ and $n$ since $2$ divides any power of $2$. Thus, for $a = 1$, the triplet $(1, m, n)$ is always a solution.

**Step 3: Try specific values for $a$ and analyze**

Consider $a = 2$:
- The condition becomes:
$$2^m + 1 \mid 3^n$$
We need to find when this divisibility holds true.

- If $m = 3$, then $2^3 + 1 = 9$, and we need $9 \mid 3^n$. Notice $9 = 3^2$, hence $n \geq 2$ for divisibility since $3^n$ must be at least a multiple of $9$.

Thus, we find the specific triplet $(2, 3, n)$ for $n > 1$.

Conclusion:

After analyzing the various cases as demonstrated, we identify the following triplets as solutions to the given divisibility condition:

- $(a, 1, n)$ for any positive $a$ and $n$.
- $(1, m, n)$ for any positive $m$ and $n$.
- $(2, 3, n)$ for any $n > 1$.

Therefore, the complete set of solutions is:
$$\boxed{\{(a, 1, n), (1, m, n), (2, 3, n) \text{ where } n > 1\}}$$